SNC12 (k) → Sn (k) + Cl2 (g) 2 SnCl2 (k) + 2 Cl2 (g) → 2 SnCl4 (s) +325.1 kJ -372.4 kJ V un agagmn --..l 9- Sn (k) + 2 Cl2 (g) → SnCl4 (s) AH =? a) 47.3 kJ O B) -511.3 kJ O NS) +697.5kJ OD) +511.3 k TO) -697.5kJ
In the given question we have to calculate the overall enthalpy of the chemical reaction.
we have to use Hess's law which states that the change in enthalpy will be equal to the sum of enthalpies individual chemical reaction.
i.e ∆Hrxn = H1 + H2
Using Hess' Law, we can use the other two reaction equations are given.
SnCl2(s) -------> Sn(s) + Cl2(g) ∆H = +325.1 kJ (equation-1 )
by reversing the equation-1 and multiply 2 both side,
here by reversing equation-1 enthalpy become negative.
we get,
2 Sn(s) + 2 Cl2 (g) --------> 2 SnCl2(s) ∆H = 2 × -325.1 kJ (equation - 2)
now
2 SnCl2(s) + 2 Cl2(g) ------> 2 SnCl4(l) ∆H = -372.4 kJ (equation - 3)
Adding (equation-2 and equation-3) them we get,
2 Sn(s) + 2 Cl2(g) + 2 SnCl2(s) + 2 Cl2(g) -------> 2 SnCl2(s) + 2 SnCl4(l) (equation - 4)
∆H = 2× -325.1 kJ + (-372.4) kJ (added the ∆H values)
Canceling common species from equation-4 on both side
2 Sn(s) + 4 Cl2(g) -----> 2 SnCl4(l) (equation-5)
now, change in enthalpy for 2 mol each species will be
∆H = 2× -325.1 + (-372.4)
= - 650.2 + (-372.4)
= -1022.6 kJ
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